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Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1

by gibguitar
Tags: curve, equation, line, slope, tangent, y1 or x1
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gibguitar
#1
Feb20-12, 03:45 PM
P: 9
1. The problem statement, all variables and given/known data

Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1

2. Relevant equations

y=1/x-1

3. The attempt at a solution

Slope of -1 means y=-1x+k

So...
-1x+k = 1/x-1

I don't know how to rearrange this into a quadratic equation so that I can solve this.
I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
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Mark44
#2
Feb20-12, 04:17 PM
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P: 21,215
Quote Quote by gibguitar View Post
1. The problem statement, all variables and given/known data

Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1

2. Relevant equations

y=1/x-1

3. The attempt at a solution

Slope of -1 means y=-1x+k

So...
-1x+k = 1/x-1

I don't know how to rearrange this into a quadratic equation so that I can solve this.
I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
You're missing a very important piece: how to find the slope of the tangent line to y = 1/x - 1.

I'm going to guess that you are in a class that has discussed how to find the tangent to a curve.
Mark44
#3
Feb20-12, 04:26 PM
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P: 21,215
OTOH, maybe this actually is a precalculus-type problem. What does it mean to say that a line is tangent to a curve?

For your equation, -x + k = 1/x - 1, what about multiplying both sides by x?

gibguitar
#4
Feb20-12, 04:34 PM
P: 9
Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1

Quote Quote by Mark44 View Post
OTOH, maybe this actually is a precalculus-type problem. What does it mean to say that a line is tangent to a curve?

For your equation, -x + k = 1/x - 1, what about multiplying both sides by x?
(-x+k)(x) = -x^2+k(x)
(1/x-1)(x) = ?

You're right, but I'm confused as to how to multiply the second part? x/x^2-x ?
gibguitar
#5
Feb20-12, 04:45 PM
P: 9
Here's my second stab at it...

-1x+k = 1/x-1

Rearranged: -x^2+kx-1

Using the discriminant of the quadratic formula:

(b^2-4ac)

a=-1 b=k^2 c=-1

k^2-4(-1)(-1)=0
Solving:
k = -2

Therefore the equation is y=-1x-2
For the line with a slope of -1 that is tangent to the curve of 1/x-1
Can anybody verify if I got this right?
Mark44
#6
Feb20-12, 04:49 PM
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P: 21,215
Quote Quote by gibguitar View Post
Here's my second stab at it...

-1x+k = 1/x-1

Rearranged: -x^2+kx-1
Where did the = go? Also, you have an error.
Quote Quote by gibguitar View Post

Using the discriminant of the quadratic formula:

(b^2-4ac)

a=-1 b=k^2 c=-1

k^2-4(-1)(-1)
Solving:
k = -2

Therefore the equation is y=-1x-2
For the line with a slope of -1 that is tangent to the curve of 1/x-1
Can anybody verify if I got this right?
Mark44
#7
Feb20-12, 04:50 PM
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P: 21,215
Quote Quote by gibguitar View Post
(-x+k)(x) = -x^2+k(x)
(1/x-1)(x) = ?

You're right, but I'm confused as to how to multiply the second part? x/x^2-x ?
What is x * (1/x)?
What is x * (-1)?
Mark44
#8
Feb20-12, 04:52 PM
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P: 21,215
From post #3:
What does it mean to say that a line is tangent to a curve?
Ray Vickson
#9
Feb20-12, 08:33 PM
Sci Advisor
HW Helper
Thanks
P: 4,958
Quote Quote by gibguitar View Post
1. The problem statement, all variables and given/known data

Find the equation of the line with slope -1 that is tangent to the curve y=1/x-1

2. Relevant equations

y=1/x-1

3. The attempt at a solution

Slope of -1 means y=-1x+k

So...
-1x+k = 1/x-1

I don't know how to rearrange this into a quadratic equation so that I can solve this.
I know I use b^2-4ac=0 and substituting for a, b and c.... I'm just stuck on rearranging the equation into the form: ax^2+bx+c=0
What you wrote means [itex] (1/x) - 1.[/itex] Is that what you meant, or did you really mean [itex] 1/(x-1)[/itex]? If you meant the latter, you need to use brackets, but if you meant the former then what you wrote is OK.

RGV
Mark44
#10
Feb20-12, 08:39 PM
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P: 21,215
Quote Quote by Ray Vickson View Post
What you wrote means [itex] (1/x) - 1.[/itex] Is that what you meant, or did you really mean [itex] 1/(x-1)[/itex]? If you meant the latter, you need to use brackets, but if you meant the former then what you wrote is OK.
I think he meant (1/x) - 1, but I'm not certain of it.


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